Question

Given that z is a standard normal random variable, find z for each situation (to 2 decimals)

a. the area to the left of z is 0.9732

b. the area between 0 and z is 0.4732

c. the area to the left of z is 0.8643

d. the area to the right of z is 0.1251

e. the are to the left of z is 0.6915

f. the are to the right of z is 0.3085

Answer 1

Solution:
a. THe area to the left of z is 0.9732 => 1.93

b. The area between o and Z is 0.4732 => 1.93
  
P(0 < Z < 1.93) = 0.4732

c. The area to the left of z is 0.8643 => 1.099

P (Z < 1.099) = 0.8643

d. The area to the right of z is 0.1251 => 1.1499

P(Z > 1.1499) = 1 − P(Z < 1.1499) = 1 − 0.8749 = 0.1251

e. The are to the left of z is 0.6915 => 0.5

P(Z < 0.5) = 0.6915

f. The are to the right of z is 0.3085 => 0.5

P(Z > 0.5) = 1 − PZ < 0.5) = 1 − 0.6915 = 0.3085